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Date May 2014 Marks available 2 Reference code 14M.2.sl.TZ2.1
Level SL only Paper 2 Time zone TZ2
Command term Calculate Question number 1 Adapted from N/A

Question

Tomek is attending a conference in Singapore. He has both trousers and shorts to wear. He also has the choice of wearing a tie or not.

The probability Tomek wears trousers is \(0.3\). If he wears trousers, the probability that he wears a tie is \(0.8\).

If Tomek wears shorts, the probability that he wears a tie is \(0.15\).

The following tree diagram shows the probabilities for Tomek’s clothing options at the conference.


Find the value of

(i)     \({\text{A}}\);

(ii)     \({\text{B}}\);

(iii)     \({\text{C}}\).

[3]
a.

Calculate the probability that Tomek wears

(i)     shorts and no tie;

(ii)     no tie;

(iii)     shorts given that he is not wearing a tie.

[8]
b.

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on both days.

[2]
c.

The conference lasts for two days.

Calculate the probability that Tomek wears trousers on one of the days, and shorts on the other day.

[3]
d.

Markscheme

(i)     \(0.7 \left( {\frac{{70}}{{100}},{\text{ }}\frac{7}{{10}},{\text{ 70% }}} \right)\)     (A1)

(ii)     \(0.2 \left( {\frac{{20}}{{100}},{\text{ }}\frac{2}{{10}},{\text{ }}\frac{1}{5},{\text{ 20% }}} \right)\)     (A1)

(iii)     \(0.85 \left( {\frac{{85}}{{100}},{\text{ }}\frac{{17}}{{20}},{\text{ 85% }}} \right)\)     (A1)

[3 marks]

a.

(i)     \(0.7 \times 0.85\)     (M1)

 

Note: Award (M1) for multiplying their values from parts (a)(i) and (a)(iii).

 

\( = 0.595{\text{ }}\left( {\frac{{119}}{{200}},{\text{ 59.5% }}} \right)\)     (A1)(ft)(G1)

 

Note: Follow through from part (a).

 

(ii)     \(0.3 \times 0.2 + 0.7 \times 0.85\)     (M1)(M1)

 

Note: Award (M1) for their two products, (M1) for adding their two products.

 

\( = 0.655{\text{ }}\left( {\frac{{131}}{{200}},{\text{ 65.5% }}} \right)\)     (A1)(ft)(G2)

 

Note: Follow through from part (a).

 

(iii)     \(\frac{{0.595}}{{0.655}}\)     (A1)(ft)(A1)(ft)

 

Notes: Award (A1)(ft) for correct numerator, (A1)(ft) for correct denominator. Follow through from parts (b)(i) and (ii).

 

\( = 0.908{\text{ }}\left( {{\text{0.90839}} \ldots ,{\text{ }}\frac{{119}}{{131}},{\text{ 90,8% }}} \right)\)     (A1)(ft)(G2)

[8 marks]

b.

\(0.3 \times 0.3\)     (M1)

\( = 0.09 \left( {\frac{9}{{100}}, 9\%} \right)\)     (A1)(G2)

[2 marks]

c.

\(0.3 \times 0.7\)     (M1)

\(0.3 \times 0.7 \times 2\)   OR   \((0.3 \times 0.7) + (0.7 \times 0.3)\)     (M1)

 

Note: Award (M1) for their correct product seen, (M1) for multiplying their product by 2 or for adding their products twice.

 

\( = 0.42 \left( {\frac{{42}}{{100}}, \frac{{21}}{{50}}, 42\%} \right)\)     (A1)(ft)(G2)

 

Note: Follow through from part (a)(i).

 

[3 marks]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.

Syllabus sections

Topic 3 - Logic, sets and probability » 3.7 » Probability of combined events, mutually exclusive events, independent events.
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